The latter product is the well-known Eulerian sine product[1]} whereas the former product has to be \(\zeta \) -regularized. By taking the Logarithm of the product one obtains formally:
\(\log {\left(\prod _{n=1}^{+\infty }\frac{i\pi }{\alpha n^2}\right)}=\log {\left(\frac{i\pi }{\alpha }\right)}\sum _{n=1}^{+\infty }1-2\sum _{n=1}^{+\infty }\log {n}\) 
